`(dr)/(ds) = 0.75r` Find the general solution of the differential equation

The general solution of a differential equation in a form of  can be evaluated using direct integration. The derivative of y denoted as `y'` can be written as `(dy)/(dx)` then `y'= f(x) ` can be expressed as `(dy)/(dx)= f(x)`

For the problem `(dr)/(ds)=0.75r` , we may apply variable separable differential equation in which we set it up as `f(y) dy= f(x) dx` .

Then,`(dr)/(ds)=0.75r` can be rearrange into `(dr)/r=0.75 ds` .

Applying direct integration on both sides:

`int...

The general solution of a differential equation in a form of  can be evaluated using direct integration. The derivative of y denoted as `y'` can be written as `(dy)/(dx)` then `y'= f(x) ` can be expressed as `(dy)/(dx)= f(x)`

For the problem `(dr)/(ds)=0.75r` , we may apply variable separable differential equation in which we set it up as `f(y) dy= f(x) dx` .

Then,`(dr)/(ds)=0.75r` can be rearrange into `(dr)/r=0.75 ds` .

Applying direct integration on both sides:

`int (dr)/r= int 0.75 ds` .

For the left side, we apply the basic integration formula for logarithm: `int (du)/u = ln|u|+C`

`int (dr) /r = ln|r|`

For the right side, we may apply the basic integration property: `int c*f(x)dx= c int f(x) dx` .

`int 0.75 ds=0.75int ds` .

Then the indefinite integral will be:

`0.75int ds= 0.75s+C`

Combining the results for the general solution of differential equation: 

`ln|r|=0.75s+C`

 `r= Ce^(0.75s)`